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Proof step 15 from proof inv-recover (previous step)

remove-refsize0(gstore1) = gstore1, remove-refsize0(istore1) = istore1, elems nx ∩ ginval2 = ∅, dom gstore1 ∪ elems nx = dom gstore2 ∪ ginval2, btree-inv(rp, adr2, istore2, istore1, zh, ri, fi), inv(crash(gstore2, gstore1, ginval2, nx) \ ginval2), inv(gstore2 \ ginval2), inv(gstore1), inv(istore2), inv(istore1), istore1 ⊑ istore2, pstate1.journal? → nx ≠ [] ∧ pstate1.lnum = nx.last ∧ pstate1.lnum ∈ gstore2, pstate1.index? → pstate1.lnum ∈ istore2 ⊦ dups(nx), ∃ ri, fi. btree-inv(null, adr2, remove-refsize0(crash(istore2, istore1)), istore1, zh0, ri, fi)

User applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: fi, fi abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ istore0 ⊑ istore ∧ remove-refsize0(istore0) = istore0 ∧ btree-inv(rp, adr, istore, istore0, zh0, ri, fi) ∧ inv(istore) → btree-inv (null, adr, remove-refsize0(crash(istore, istore0)), istore0, zh1, fi, fi); (ss) Used terms: fi, fi